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To product a product $3$ square plates $A$ with measures $300\times 500$mm and $2$ square plates $B$ with measures $400\times 600$mm are needed. These plates must be cut out by two different workpieces with measres $1000\times 1000$mm (Type I) and $600\times 1500$mm (Type II). The plates, for technology reason, must be cut through and no just cut in.

Totally $500$ products must be produced. The costs for the workpieces are proportional to their area, i.e., if a workpiece of Type I has the price 1, then a workpiece of Type II has the price $0.9$ ($1000000\ mm^2$ to $900000\ mm^2$).

There are $4$ possibilities, to cut plates A and/or B from a workpiece of Type I:

(1) 4A (scrap $16\%$)
(2) 3A, 1B (scrap $13\%$)
(3) 1A, 2B (scrap $31\%$)
(4) 3B (scrap $28\%$)

There are $3$ possibilities, to cut plates A and/or B from a workpiece of Type II:

(5) 4A ($8\%$ scrap)
(6) 2A, 2B (scrap $0\%$)
(7) 3B (scrap $24\%$)

Try to find these possibilities.

Formulate the problem as a linear optimization problem. Let $x_1, x_2, \ldots , x_7$ be the number of workpieces, that are necessary for the schemes (1), (2), ... , (7). The costs for the workpieces must be minimized.

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What does it mean to find the possibilities? What do we have to do? Could you give me a hint?

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    Please check the numbers. In my onpinion it could be "... plates $A$ with measures $300\times\color{red}{700}$mm and..." Then the scrap can be calculated: **Type I, 3A, 1B**: $1-\frac{3*300*700+1*400*600}{1000*1000}=0.13=13\%$. But if I´m right it doesn´t work for some Type II combinations. Please check it by yourself. – 2017-02-17
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    Maybe we shouldn´t care about how the scrap has been evaluated. What do you think ? – 2017-02-17
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    I got stuck because I haven't really understood what we are supposed to do. Do we want to confirm that these 7 schemes are possible? @callculus – 2017-02-17
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    And for the linear optimization problem these 7 schemes are the constraints? @callculus – 2017-02-17
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    I don´t think so. It is assumed they are possible only. In general you have to minimize the scrap by using the numbers in the brackets. I have an Idea but it is not formulated in a linear program yet. Reagarding your last comment, the 7 "conditions" have to be used to formulate the objective function-I think. But they have to be used for the constraints as well – 2017-02-17
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    For the objective function, since we want to minimize the cost, we have to use also the information that the costs for the workpieces are proportional to their area, right? @callculus – 2017-02-17
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    Yes, I think that too. For instance with (1) you can produce 4A with a scrap of $ 0.16$ and a (relative) price of $1$. With (5) you can produce 4A as well with a scrap of $0.08$ and a (relative) price of $0.9$ If I´m right then you need (at least) 1500 A-plates and 1000 B-plates. So one condition is $4x_1+3x_2+x_3+4x_5+2x_6\geq 1500$. But I have to think for a while till I can be more specific. – 2017-02-17
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    For one product you need 3 plates A. And you have to produce 500 products. : $3\cdot 500=1500$. Similar for the B-plates. I think the objective function is just $0.16\cdot 1\cdot x_1+0.13\cdot 1\cdot x_2 +0.31\cdot 1\cdot x_3+0.28\cdot 1\cdot x_4+0.08 \cdot 0.9\cdot x_5+0 \cdot 0.9\cdot x_6+0.24 \cdot 0.9\cdot x_7$ – 2017-02-17
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    So, we multiply every $x_i$ by the percentage of the scrap and by 1 for $i\leq 4$ and by 0 for $4\leq i\leq 7$, right? We multiply by the percentage of scrap to minimize the scrap, right? Also, we have just these two conditions, right? @callculus – 2017-02-17
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    Yes. if you mean 0.9 instead of 0. That´s a good question. But I´m not sure if we do not need other conditions additionally. Maybe you find some in the next few hours. Write down for yourself the parts of the model you have so far (and compare it with the text). Then you have a better overview if something is missing or not. – 2017-02-17
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    To minimize the cost for the workpieces is equivalent to minimize the scrap? @callculus – 2017-02-17
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    That´s another good question. One of my thought was that we should incorporate the scrap numbers into the model. Another thought was that (the cost of) the scrap has to be minimized. That should be the goal. If this is equivalent to the minimizing of the cost for the workpieces I cannot say for sure. I´m off for a while. – 2017-02-17
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    Ok. Thank you very much for your help!! :-) @callculus – 2017-02-17

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